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Is there an example of a closed manifold $M$ with a proper subset $A\subset M$ such the inclusion $i:A \to M$ gives a ring isomorphism $i^{*}$ between $\mathbb{Z}$-cohomologies?

In this question $A$ is merely a proper subset.(not necessarily compact, not necessarily submanifold)

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$*\to\mathbb{R}$ – Adam Hughes Jan 30 '14 at 22:45
@Adam Note that a closed manifold (as usual) is a compact manifold without boundary! – Ali Taghavi Jan 30 '14 at 22:47
up vote 7 down vote accepted

Suppose for simplicity that $M$ is connected and orientable. Choose a point $x\in M\setminus A$, and a closed disc $U$ centred at $x$. We can use the chart to deform $i$ into a homotopic map $j$ such that $j(A)\subseteq M\setminus\text{int}(U)$. Collapsing the complement of $U$ gives a map $p$ from $M$ to the one-point compactification $U\cup\{\infty\}$, which is homeomorphic to $S^n$. It is standard that the resulting map $p^*\colon\mathbb{Z}=\widetilde{H}^n(S^n)\to H^n(M)$ is an isomorphism, but $pj$ is constant, so the map $i^*p^*=j^*p^*=(pj)^*$ is zero, so $i^*$ is not an isomorphism.

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I did not assume that $A$ is closed(It is possible that $A$ is dense. so we can not choose the open disc $U$. – Ali Taghavi Jan 30 '14 at 22:58
This does not make much difference. I have modified the argument. – Neil Strickland Jan 30 '14 at 23:22
In the non orientable case, a similar argument shows that the induced homomorphism $H_n(A;\mathbb{Z}_2)\to H_n(M;\mathbb{Z}_2)$ is trivial. On the other hand, if the inclusion induced an isomorphism with $\mathbb{Z}$ coefficients, it would also be an isomorphism with any coefficients (universal coefficient formula). – Alex Degtyarev Jan 31 '14 at 5:29

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